Evaluate: `int1/(sinx(2cos^(2)x-1))dx`

1.	Evaluate: `int1/(sinx(2cos^(2)x-1))dx`
Answer» Putting `cosx=t`, we get `I=int(1/(sinx(2cos^(2)x-1)))dx=int(1/(sinx(2t^(2)-1))xx -(dt)/(sinx))=-int1/((1-t^(2)(2t^(2)-1))`dt `therefore I=-int(1/(1-t^(2))+2/(2t^(2)-1))dt=-int1/(1-t^(2))dt-2int1/(2t^(2)-1)dt` `=-1/2ln \|(1+t)/(1-t)\|-sqrt(2)/2ln \|(sqrt(2)t-1)/(sqrt(2)t+1)\|+C=-1/2ln\|(1+cosx)/(1-cosx)\|-1/sqrt(2)ln\|(sqrt(2)cosx-1)/(sqrt(2)cosx+1)\|+C`

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Evaluate: `int1/(sinx(2cos^(2)x-1))dx`

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